已知:xyz=1,x+y+z=2,x^2+y^2+z^2=16求1/(xy+2z)+1/(zy+2x)+1/(xz+2y)的值
问题描述:
已知:xyz=1,x+y+z=2,x^2+y^2+z^2=16求1/(xy+2z)+1/(zy+2x)+1/(xz+2y)的值
答
xy+2z=xy+4-2x-2y=(x-2)(y-2).同理,yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),xy+yz+zx=-6.(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.原式=[(x-2)+(y-2)+(z-2)]...